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3 years ago

Which expression is equivalent to (2mn)^4/6m^-3n^-2? Assume .m=0,n=0

Mathematics

2 answers:

Elza [17]3 years ago

7 0

Given expression: $\frac{(2mn)^4}{6m^{-3}n^{-2}}$ .

$\mathrm{Apply\:exponent\:rule}:\quad \left(ab\right)^c=a^cb^c$

$\left(2mn\right)^4:\quad 2^4m^4n^4$

$=\frac{2^4m^4n^4}{6m^{-3}n^{-2}}$

$=\frac{2^4m^4n^4}{2\cdot \:3m^{-3}n^{-2}}$

$=\frac{2^3m^4n^4}{3m^{-3}n^{-2}}$

$\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}\:=\:x^{a-b}$

$\frac{m^4}{m^{-3}}=m^{4-\left(-3\right)}=m^7$

$=\frac{2^3m^7n^4}{3n^{-2}}$

$\frac{n^4}{n^{-2}}=n^{4-\left(-2\right)}=n^6$

$=\frac{2^3m^7n^6}{3}$

$=\frac{8m^7n^6}{3}\$

<h3>Therefore, correct option is first option $\frac{8m^7n^6}{3}.$ </h3>

kati45 [8]3 years ago

4 0

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Math! 30 points + brainliest

Ronch [10]

Answer:

A) 9.56x10^38 ergs

B) 7.4x10^-3 mm

Step-by-step explanation:

A) 9.56x10^38 ergs B) 7.4x10^-3 mm A). For the sun, just multiply the power by time, so 3.9x10^33 erg/sec * 2.45x10^5 sec = 9.56x10^38 B) Of the two values 7.4x10^-3 and 7.4x10^3, the value 7.4x10^-3 is far more reasonable as a measurement for blood cell. Reason becomes quite evident if you take the 7.4x10^3 value and convert to a non-scientific notation value. Since the exponent is positive, shift the decimal point to the right. So 7.4x10^3 mm = 7400 mm, or in easier to understand terms, over 7 meters. That is way too large for a blood cell when you consider that you need a microscope to see one. Now the 7.4x10^-3 mm value converts to 0.0074 mm which is quite small and would a reasonable size for a blood cell.

6 0

3 years ago

Read 2 more answers

Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based

notsponge [240]

Answer:

The Taylor series for $sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$ , the first three non-zero terms are $9x^{2} -27x^{4}+\frac{162}{5}x^{6}$ and the interval of convergence is $( -\infty, \infty )$

Step-by-step explanation:

<u>These are the steps to find the Taylor series for the function</u> $sin^2(3 x)$

Use the trigonometric identity:

$sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$

2. The Taylor series of $cos(x)$

$cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}$

Substituting y=6x we have:

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

3. Find the Taylor series for $sin^2(3x)$

$sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$ (1)

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$ (2)

Substituting (2) in (1) we have:

$\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

Bring the factor $\frac{1}{2}$ inside the sum

$\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )$

$\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

Extract the term for n=0 from the sum:

$\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

<u>To find the first three non-zero terms you need to replace n=3 into the sum</u>

$sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}$